\(\int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx\) [202]

Optimal result

Integrand size = 28, antiderivative size = 175 \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=\frac {26 a^3 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{21 d}+\frac {26 i a^3 (e \sec (c+d x))^{5/2}}{35 d}+\frac {26 a^3 e (e \sec (c+d x))^{3/2} \sin (c+d x)}{21 d}+\frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {26 i (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}{63 d} \]

[Out]

26/35*I*a^3*(e*sec(d*x+c))^(5/2)/d+26/21*a^3*e*(e*sec(d*x+c))^(3/2)*sin(d*x+c)/d+26/21*a^3*e^2*(cos(1/2*d*x+1/
2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*(e*sec(d*x+c))^(1/2)/d
+2/9*I*a*(e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^2/d+26/63*I*(e*sec(d*x+c))^(5/2)*(a^3+I*a^3*tan(d*x+c))/d

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3579, 3567, 3853, 3856, 2720} \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=\frac {26 a^3 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{21 d}+\frac {26 i a^3 (e \sec (c+d x))^{5/2}}{35 d}+\frac {26 a^3 e \sin (c+d x) (e \sec (c+d x))^{3/2}}{21 d}+\frac {26 i \left (a^3+i a^3 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}{63 d}+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{5/2}}{9 d} \]

[In]

Int[(e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(26*a^3*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(21*d) + (((26*I)/35)*a^3*(e*Se
c[c + d*x])^(5/2))/d + (26*a^3*e*(e*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(21*d) + (((2*I)/9)*a*(e*Sec[c + d*x])^(
5/2)*(a + I*a*Tan[c + d*x])^2)/d + (((26*I)/63)*(e*Sec[c + d*x])^(5/2)*(a^3 + I*a^3*Tan[c + d*x]))/d

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3567

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3579

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*
Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] + Dist[a*((m + 2*n - 2)/(m + n - 1)), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] &&
GtQ[n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps \begin{align*} \text {integral}& = \frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {1}{9} (13 a) \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2 \, dx \\ & = \frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {26 i (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac {1}{7} \left (13 a^2\right ) \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx \\ & = \frac {26 i a^3 (e \sec (c+d x))^{5/2}}{35 d}+\frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {26 i (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac {1}{7} \left (13 a^3\right ) \int (e \sec (c+d x))^{5/2} \, dx \\ & = \frac {26 i a^3 (e \sec (c+d x))^{5/2}}{35 d}+\frac {26 a^3 e (e \sec (c+d x))^{3/2} \sin (c+d x)}{21 d}+\frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {26 i (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac {1}{21} \left (13 a^3 e^2\right ) \int \sqrt {e \sec (c+d x)} \, dx \\ & = \frac {26 i a^3 (e \sec (c+d x))^{5/2}}{35 d}+\frac {26 a^3 e (e \sec (c+d x))^{3/2} \sin (c+d x)}{21 d}+\frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {26 i (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}{63 d}+\frac {1}{21} \left (13 a^3 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {26 a^3 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{21 d}+\frac {26 i a^3 (e \sec (c+d x))^{5/2}}{35 d}+\frac {26 a^3 e (e \sec (c+d x))^{3/2} \sin (c+d x)}{21 d}+\frac {2 i a (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^2}{9 d}+\frac {26 i (e \sec (c+d x))^{5/2} \left (a^3+i a^3 \tan (c+d x)\right )}{63 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.74 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.51 \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=\frac {a^3 \sec ^2(c+d x) (e \sec (c+d x))^{5/2} \left (728 i+1008 i \cos (2 (c+d x))+1560 \cos ^{\frac {9}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-150 \sin (2 (c+d x))+195 \sin (4 (c+d x))\right )}{1260 d} \]

[In]

Integrate[(e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(a^3*Sec[c + d*x]^2*(e*Sec[c + d*x])^(5/2)*(728*I + (1008*I)*Cos[2*(c + d*x)] + 1560*Cos[c + d*x]^(9/2)*Ellipt
icF[(c + d*x)/2, 2] - 150*Sin[2*(c + d*x)] + 195*Sin[4*(c + d*x)]))/(1260*d)

Maple [A] (verified)

Time = 7.50 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.05

\[-\frac {2 i e^{2} a^{3} \sqrt {e \sec \left (d x +c \right )}\, \left (195 F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+195 F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+195 i \tan \left (d x +c \right )-252 \left (\sec ^{2}\left (d x +c \right )\right )-135 i \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )+35 \left (\sec ^{4}\left (d x +c \right )\right )\right )}{315 d}\]

[In]

int((e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^3,x)

[Out]

-2/315*I*e^2*a^3/d*(e*sec(d*x+c))^(1/2)*(195*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*(
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+195*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2
)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+195*I*tan(d*x+c)-252*sec(d*x+c)^2-135*I*sec(d*x+c)^2*tan(d*x+c)+35*sec(d*x
+c)^4)

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.08 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.47 \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (195 i \, a^{3} e^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 1158 i \, a^{3} e^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 1456 i \, a^{3} e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 858 i \, a^{3} e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 195 i \, a^{3} e^{2}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 195 \, \sqrt {2} {\left (i \, a^{3} e^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 i \, a^{3} e^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 i \, a^{3} e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, a^{3} e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} e^{2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{315 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate((e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-2/315*(sqrt(2)*(195*I*a^3*e^2*e^(8*I*d*x + 8*I*c) - 1158*I*a^3*e^2*e^(6*I*d*x + 6*I*c) - 1456*I*a^3*e^2*e^(4*
I*d*x + 4*I*c) - 858*I*a^3*e^2*e^(2*I*d*x + 2*I*c) - 195*I*a^3*e^2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I
*d*x + 1/2*I*c) + 195*sqrt(2)*(I*a^3*e^2*e^(8*I*d*x + 8*I*c) + 4*I*a^3*e^2*e^(6*I*d*x + 6*I*c) + 6*I*a^3*e^2*e
^(4*I*d*x + 4*I*c) + 4*I*a^3*e^2*e^(2*I*d*x + 2*I*c) + I*a^3*e^2)*sqrt(e)*weierstrassPInverse(-4, 0, e^(I*d*x
+ I*c)))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c)
+ d)

Sympy [F]

\[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=- i a^{3} \left (\int i \left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx + \int \left (- 3 \left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \tan {\left (c + d x \right )}\right )\, dx + \int \left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \tan ^{3}{\left (c + d x \right )}\, dx + \int \left (- 3 i \left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \tan ^{2}{\left (c + d x \right )}\right )\, dx\right ) \]

[In]

integrate((e*sec(d*x+c))**(5/2)*(a+I*a*tan(d*x+c))**3,x)

[Out]

-I*a**3*(Integral(I*(e*sec(c + d*x))**(5/2), x) + Integral(-3*(e*sec(c + d*x))**(5/2)*tan(c + d*x), x) + Integ
ral((e*sec(c + d*x))**(5/2)*tan(c + d*x)**3, x) + Integral(-3*I*(e*sec(c + d*x))**(5/2)*tan(c + d*x)**2, x))

Maxima [F]

\[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \,d x } \]

[In]

integrate((e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((e*sec(d*x + c))^(5/2)*(I*a*tan(d*x + c) + a)^3, x)

Giac [F]

\[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \,d x } \]

[In]

integrate((e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(5/2)*(I*a*tan(d*x + c) + a)^3, x)

Mupad [F(-1)]

Timed out. \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^3 \, dx=\int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3 \,d x \]

[In]

int((e/cos(c + d*x))^(5/2)*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

int((e/cos(c + d*x))^(5/2)*(a + a*tan(c + d*x)*1i)^3, x)